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Question Number 192112 by Spillover last updated on 08/May/23

$$provethat.0!=1$$

Commented by Frix last updated on 08/May/23

$$\mathrm{I}\mathrm{think}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{defined}0!=1$$$${\mathrm{There}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{idea}\mathrm{of}\mathrm{the}‘‘\mathrm{Empty}{\mathrm{Product}}^{″}$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{obvious}\mathrm{that}\mathrm{the}‘‘\mathrm{Empty}{\mathrm{Sum}}^{″}=0$$$$\mathrm{because}\mathrm{if}\mathrm{we}\mathrm{sum}\mathrm{up}\mathrm{nothing}\mathrm{we}\mathrm{still}$$$$\mathrm{have}\mathrm{nothing}.\mathrm{Because}\mathrm{the}\mathrm{sum}$$$$\log x+\log y=\log xy\mathrm{we}\mathrm{set}\mathrm{the}‘‘\mathrm{Empty}$$$${\mathrm{Product}}^{″}=1$$$$x!=\underset{k=1}{\prod ^x}\Rightarrow 0!\mathrm{is}\mathrm{the}‘‘\mathrm{Empty}{\mathrm{Product}}^{″}$$$$$$$$$$$$\mathrm{But}\mathrm{we}\mathrm{can}\mathrm{simply}\mathrm{calculate}\mathrm{backwards}$$$$3!=6$$$$2!=\frac{3!}{3}=2$$$$1!=\frac{2!}{2}=1$$$$0!=\frac{1!}{1}=1$$$$[\Rightarrow x!\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x\in {\mathbb{Z}}^{-}]$$

Commented by Spillover last updated on 12/May/23

$$thanks$$

Answered by Skabetix last updated on 08/May/23

$$Hello$$$$X!=X(X-1)!$$$$\iff (X-1)!=\frac{X!}{X}$$$$Nowwejusthavetotestfor0$$$$(1-1)!=\frac{1!}{1}$$$$\iff 0!=\frac{1}{1}=1$$$$$$

Answered by Spillover last updated on 12/May/23

$${}^n{P}_r=\frac{n!}{(n-r)!}$$$${}^n{P}_n=\frac{n!}{(n-n)!}=\frac{n!}{\left(0\right)!}$$$$but{}^n{P}_n=n!$$$$n!=\frac{n!}{0!}0!=\frac{n!}{n!}=1$$$$0!=1$$$$$$

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